import sun.nio.cs.FastCharsetProvider;

import java.util.*;

public class BinaryTree {

    static class TreeNode {
        public char val;
        public TreeNode left;//左孩子的引用
        public TreeNode right;//右孩子的引用

        public TreeNode(char val) {
            this.val = val;
        }
    }


    /**
     * 创建一棵二叉树 返回这棵树的根节点
     *
     * @return
     */
    public TreeNode createTree() {
        TreeNode tA = new TreeNode('A');
        TreeNode tB = new TreeNode('B');
        TreeNode tC = new TreeNode('C');
        TreeNode tD = new TreeNode('D');
        TreeNode tE = new TreeNode('E');
        //TreeNode tF = new TreeNode('F');
        tA.left = tB;
        tA.right = tC;
        tB.left = tD;
        tB.right = tE;
        //tD.left  = tF;
        return  tA;
    }

    // 前序遍历
    public void preOrder(TreeNode root) {
        if(root == null) {
            return;
        }
        System.out.print(root.val + " ");
        preOrder(root.left);
        preOrder(root.right);
    }

    // 中序遍历
    void inOrder(TreeNode root) {
        if(root == null) {
            return;
        }
        preOrder(root.left);
        System.out.print(root.val + " ");
        preOrder(root.right);
    }

    // 后序遍历
    void postOrder(TreeNode root) {
        if(root == null) {
            return;
        }
        preOrder(root.left);
        preOrder(root.right);
        System.out.print(root.val + " ");
    }

    public static int nodeSize;

    /**
     * 获取树中节点的个数：遍历思路
     */
    public void size(TreeNode root) {
        if(root == null) {
            return;
        }
        nodeSize++;
        size(root.left);
        size(root.right);
    }

    /**
     * 获取节点的个数：子问题的思路
     *
     * @param root
     * @return
     */
    int size2(TreeNode root) {
        if (root == null) {
            return  0;
        }
        return size2(root.left) + size2(root.right) + 1;
    }


    /*
     获取叶子节点的个数：遍历思路
     */
    public static int leafSize = 0;

    void getLeafNodeCount1(TreeNode root) {
        if(root == null) {
            return;
        }
        if(root.left == null && root.right == null) {
            leafSize++;
        }
        getLeafNodeCount1(root.left);
        getLeafNodeCount1(root.right);
    }

    /*
     获取叶子节点的个数：子问题
     */
    int getLeafNodeCount2(TreeNode root) {
        if(root == null) {
            return 0;
        }
        if(root.left == null && root.right == null) {
            return 1;
        }
        return getLeafNodeCount2(root.left) + getLeafNodeCount2(root.right);
    }

    /*
    获取第K层节点的个数
     */
    int getKLevelNodeCount(TreeNode root, int k) {
        if (root == null) {
            return 0;
        }
        if (k==1) {
            return 1;
        }
        return getKLevelNodeCount(root.left,k-1) + getKLevelNodeCount(root.right,k-1);
    }

    /*
     获取二叉树的高度
     时间复杂度：O(N)
     */
    int getHeight(TreeNode root) {
        if(root == null) {
            return 0;
        }
        if(root.left == null && root.right ==null){
            return 1;
        }
        int leftHigh = getHeight(root.left);
        int rightHigh = getHeight(root.right);

        return leftHigh > rightHigh ? leftHigh + 1 : rightHigh + 1;
    }


    // 检测值为value的元素是否存在
    TreeNode find(TreeNode root, char val) {
        if(root == null) {
            return null;
        }
        if(root.val == val) {
            return root;
        }
        TreeNode nodeLeft = find(root.left,val);
        TreeNode nodeRight = find(root.right,val);

        return nodeLeft == null ? nodeRight : nodeLeft;
    }

    //层序遍历
    void levelOrder(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<>();
        if(root != null) {
            queue.offer(root);
            while(!queue.isEmpty()) {
                TreeNode cur = queue.poll();
                System.out.print(cur.val + " ");
                if(cur.left != null) {
                    queue.offer(cur.left);
                }
                if (cur.right != null) {
                    queue.offer(cur.right);
                }
            }
        }
        System.out.println();
    }

    List<List<TreeNode>> levelOrder2(TreeNode root) {
        List<List<TreeNode>> ret = new ArrayList<>();
        if(root == null) {
            return ret;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        int j = 0;
        while(!queue.isEmpty()) {
            int size = queue.size();
            List<TreeNode> temp = new ArrayList<>();
            for(int i = 0 ; i < size; ++i) {
                TreeNode cur = queue.poll();
                temp.add(cur);
                if(cur.left != null) {
                    queue.offer(cur.left);
                }
                if (cur.right != null) {
                    queue.offer(cur.right);
                }
            }
            ret.add(temp);
        }
        return ret;
    }

    // 判断一棵树是不是完全二叉树
    boolean isCompleteTree(TreeNode root) {
        Deque<TreeNode> dque = new LinkedList<>();
        boolean ret = true;
        dque.offer(root);
        while (!dque.isEmpty()) {
            int n = dque.size();
            for(int i = 0; i < n; ++i) {
                TreeNode cur = dque.poll();
                if(cur == null) {
                    ret = false; //第一次遇到null变为false;
                }
                else {
                    if(!ret) {  //上一次遇到null,这次不是null，说明当前节点存在不是完全二叉树
                        return false;
                    }
                    dque.offer(cur.left);
                    dque.offer(cur.right);
                }
            }

        }
        return true;
    }
}
